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Meyerhofs Method

Courses > Foundation Analysis and Design > Bearing Capacity of Shallow Foundations > Meyerhofs Method


Introduction on Meyerhof's Method :

In geotechnical engineering, bearing capacity is the capacity of soil to support the loads applied to the ground. The bearing capacity of soil is the maximum average contact pressure between the foundation and the soil which should not produce shear failure in the soil. Ultimate bearing capacity (qf) is the theoretical maximum pressure which can be supported without failure; allowable bearing capacity (qa) is the ultimate bearing capacity divided by a factor of safety. Sometimes, on soft soil sites, large settlements may occur under loaded foundations without actual shear failure occurring; in such cases, the allowable bearing capacity is based on the maximum allowable settlement.

There are three modes of failure that limit bearing capacity: general shear failure, local shear failure, and punching shear failure.

Note! Other methods proposed for bearing capacity of shallow foundations are: Terzaghi's bearing capacity method which is the earliest method proposed in 1943, and Brinch and Hansen (1970).

 


Concepts and Formulas of Meyerhof's Method:

Meyerhof’s general bearing capacity equations

Vertical load:

Qu = c Nc Sc Dc + g D Nq Sq Dq + 0.5 g B Ng Sg Dg

Inclined load:

Qu = c Nc Sc Dc Ic + g D Nq Sq Dq Iq + 0.5 g B Ng Sg Dg Ig

Where:

Nc, Nq, Nr:  Meyerhof’s bearing capacity factors depend on soil friction angle, f.

Nq = eptanf tan2(45+f/2)

Nc = cot f ( Nq – 1)

Ng = (Nq-1) tan (1.4f)

Sc, Sq, Sg: shape factors

Dc, Dq, Dg: depth factors

Ic, Iq, Ig: incline load factors

Friction angle

Shape factor

Depth factor

Incline load factors

Any f

Sc=1+0.2Kp(B/L)

Dc=1+0.2ÖKp (B/L)

Ic=Iq=(1-q/90°)2

f = 0

Sq=Sg=1

Dq=Dg=1

Ig=1

10°

Sq=Sg=1+0.1Kp(B/L)

Dq=Dr=1+0.1ÖKp (D/B)

Ig=(1-q/f)2

 

C: Cohesion of soil

g : unit weight of soil

D: depth of footing

B, L: width and length of footing

Kp = tan2(45+f/2), passive pressure coefficient.

q = angle of axial load to vertical axis

Table 1: Meyerhof’s bearing capacity factors

f

Nc

Nq

Nr

0

5.1

1

0

5

6.5

1.6

0.1

10

8.3

2.5

0.4

15

11

3.9

1.2

20

14.9

6.4

2.9

25

20.7

10.7

6.8

30

30.1

18.4

15.1

35

46.4

33.5

34.4

40

75.3

64.1

79.4

Figure 1: Meyerhof’s bearing capacity factors.

 

 


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Solved sample problems of Meyerhof's Method:

Example 1: Strip footing on clayey sand (English units)

Given:

3 ft wide strip footing, bottom of footing at 2 ft below ground level

Requirement

Determine allowable soil bearing capacity using Meyerhof’s equation.

Solution:

Determine ultimate soil bearing capacity using Meyerhof’s bearing capacity equation for vertical load.

Passive pressure coefficient

Kpr = tan2(45+f/2) = tan2(45+25/2) = 2.5

 Shape factors:

Sc=1+0.2Kp(B/L) = 1+0.2*2.5*(0)=1

Sq=Sg=1+0.1Kp(B/L) = 1+0.1*2.5*(0) = 1

Depth factors:

Dc=1+0.2ÖKp (B/L) = 1+0.2*Ö2 (0) = 1

Dq=Dg=1+0.1ÖKp (D/B) = 1+0.1*Ö2.5 (3/3) = 1.16

From Table 1 or Figure 1, Nc = 20.7, Nq = 10.7, Nr = 6.8 for f = 25 degree

Qu = c Nc Sc Dc + g D Nq Sq Dq + 0.5 g B Ng Sg Dg

= 500*20.7*1*1 +100*3*10.7*1*1.16+0.5*100*3*6.8*1*1.16

=  15257 lbs/ft2

Allowable soil bearing capacity,

Qa = Qu / F.S. = 15257 / 3 = 5085 lbs/ft2 @ 5000 lbs/ft2

 

Example 2: Rectangular footing on sandy clay (English units)

Given:

8 ft by 4 ft rectangular footing, bottom of footing at 3 ft below ground level.

Requirement

Determine allowable soil bearing capacity using Meyerhof’s equation.

Solution:

Determine ultimate soil bearing capacity using Meyerhof’s bearing capacity equation for vertical load.

Passive pressure coefficient

Kpr = tan2(45+f/2) = tan2(45+20/2) = 2.

Shape factors:

Sc=1+0.2Kp(B/L) = 1+0.2*2*(4/8)=1.2

Sq=Sg=1+0.1Kp(B/L) = 1+0.1*2*(4/8) = 1.1

Depth factors:

Dc=1+0.2ÖKp (B/L) = 1+0.2*Ö2 (4/8) = 1.14

Dq=Dg=1+0.1ÖKp (D/B) = 1+0.1*Ö2 (3/4) = 1.1

From Table 1 or Figure 1, Nc = 14.9, Nq = 6.4, Nr = 2.9 for f = 20 degree

Qu = c Nc Sc Dc + g D Nq Sq Dq + 0.5 g B Ng Sg Dg

= 500*14.9*1.2*1.14 +100*3*6.4*1.1*1.1+0.5*100*4*2.9*1.1*1.1

=  13217 lbs/ft2

Allowable soil bearing capacity,

Qa = Qu / F.S. = 13217 / 3 = 4406 lbs/ft2 @ 4400 lbs/ft2

 

Example 3: Square footing with incline loads (English units)

Given:

8 ft by 8 ft square footing, bottom of footing at 3 ft below ground level.

Requirement

Determine allowable soil bearing capacity using Meyerhof’s equation.

Solution:

Determine ultimate soil bearing capacity using Meyerhof’s bearing capacity equation for vertical load.

Passive pressure coefficient

Kpr = tan2(45+f/2) = tan2(45+15/2) = 1.7

Shape factors:

Sc=1+0.2Kp(B/L) = 1+0.2*1.7*(8/8)=1.34

Sq=Sg=1+0.1Kp(B/L) = 1+0.1*1.7*(8/8) =1.17

Depth factors:

Dc=1+0.2ÖKp (B/L) = 1+0.2*Ö1.7 (8/8) = 1.26

Dq=Dg=1+0.1ÖKp (D/B) = 1+0.1*Ö1.7 (3/8) = 1.05

Incline load factors:

q = tan-1 (20/100) = 11.3°

Ic=Iq=(1-q/90°)2=(1-11.3/90)2= 0.76

Ig=(1-q/f)2=(1-11.3/15)2=0.06

From Table 1 or Figure 1, Nc = 11, Nq = 3.9, Nr = 1.2 for f = 15 degree

Qu = c Nc Sc Dc Ic + g D Nq Sq Dq Iq + 0.5 g B Ng Sg Dg Ig

= 500*11*1.34*1.26*0.76+100*3*3.9*1.17*1.05*0.76+0.5*100*8*1.17*1.05*0.06

=  8179 lbs/ft2

Allowable soil bearing capacity,

Qa = Qu / F.S. = 8179 / 3 = 2726 lbs/ft2 @ 2700 lbs/ft2

 


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